What is the extraneous solution to these equations? $\dfrac{x^2 - 30}{x - 7} = \dfrac{2x + 5}{x - 7}$
Explanation: Multiply both sides by $x - 7$ $ \dfrac{x^2 - 30}{x - 7} (x - 7) = \dfrac{2x + 5}{x - 7} (x - 7)$ $ x^2 - 30 = 2x + 5$ Subtract $2x + 5$ from both sides: $ x^2 - 30 - (2x + 5) = 2x + 5 - (2x + 5)$ $ x^2 - 30 - 2x - 5 = 0$ $ x^2 - 35 - 2x = 0$ Factor the expression: $ (x + 5)(x - 7) = 0$ Therefore $x = -5$ or $x = 7$ At $x = 7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 7$, it is an extraneous solution.